3.9.64 \(\int \frac {(f+g x) (a+b x+c x^2)^{3/2}}{d+e x} \, dx\) [864]

Optimal. Leaf size=441 \[ -\frac {\left (3 b^3 e^3 g-64 c^3 d^2 (e f-d g)+16 c^2 e (5 b d-4 a e) (e f-d g)-4 b c e^2 (2 b e f-2 b d g+3 a e g)+2 c e \left (3 b^2 e^2 g+16 c^2 d (e f-d g)-4 c e (2 b e f-2 b d g+3 a e g)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 e^4}+\frac {(8 c e f-8 c d g+3 b e g+6 c e g x) \left (a+b x+c x^2\right )^{3/2}}{24 c e^2}+\frac {\left (3 b^4 e^4 g-128 c^4 d^3 (e f-d g)+192 c^3 d e (b d-a e) (e f-d g)-8 b^2 c e^3 (b e f-b d g+3 a e g)+48 c^2 e^2 \left (a^2 e^2 g-b^2 d (e f-d g)+2 a b e (e f-d g)\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} e^5}+\frac {\left (c d^2-b d e+a e^2\right )^{3/2} (e f-d g) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^5} \]

[Out]

1/24*(6*c*e*g*x+3*b*e*g-8*c*d*g+8*c*e*f)*(c*x^2+b*x+a)^(3/2)/c/e^2+1/128*(3*b^4*e^4*g-128*c^4*d^3*(-d*g+e*f)+1
92*c^3*d*e*(-a*e+b*d)*(-d*g+e*f)-8*b^2*c*e^3*(3*a*e*g-b*d*g+b*e*f)+48*c^2*e^2*(a^2*e^2*g-b^2*d*(-d*g+e*f)+2*a*
b*e*(-d*g+e*f)))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)/e^5+(a*e^2-b*d*e+c*d^2)^(3/2)*(-d*
g+e*f)*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^5-1/64*(3*b^3*e
^3*g-64*c^3*d^2*(-d*g+e*f)+16*c^2*e*(-4*a*e+5*b*d)*(-d*g+e*f)-4*b*c*e^2*(3*a*e*g-2*b*d*g+2*b*e*f)+2*c*e*(3*b^2
*e^2*g+16*c^2*d*(-d*g+e*f)-4*c*e*(3*a*e*g-2*b*d*g+2*b*e*f))*x)*(c*x^2+b*x+a)^(1/2)/c^2/e^4

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Rubi [A]
time = 0.54, antiderivative size = 441, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {828, 857, 635, 212, 738} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (48 c^2 e^2 \left (a^2 e^2 g+2 a b e (e f-d g)+b^2 (-d) (e f-d g)\right )-8 b^2 c e^3 (3 a e g-b d g+b e f)+192 c^3 d e (b d-a e) (e f-d g)+3 b^4 e^4 g-128 c^4 d^3 (e f-d g)\right )}{128 c^{5/2} e^5}-\frac {\sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e g-2 b d g+2 b e f)+3 b^2 e^2 g+16 c^2 d (e f-d g)\right )+16 c^2 e (5 b d-4 a e) (e f-d g)-4 b c e^2 (3 a e g-2 b d g+2 b e f)+3 b^3 e^3 g-64 c^3 d^2 (e f-d g)\right )}{64 c^2 e^4}+\frac {(e f-d g) \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^5}+\frac {\left (a+b x+c x^2\right )^{3/2} (3 b e g-8 c d g+8 c e f+6 c e g x)}{24 c e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x),x]

[Out]

-1/64*((3*b^3*e^3*g - 64*c^3*d^2*(e*f - d*g) + 16*c^2*e*(5*b*d - 4*a*e)*(e*f - d*g) - 4*b*c*e^2*(2*b*e*f - 2*b
*d*g + 3*a*e*g) + 2*c*e*(3*b^2*e^2*g + 16*c^2*d*(e*f - d*g) - 4*c*e*(2*b*e*f - 2*b*d*g + 3*a*e*g))*x)*Sqrt[a +
b*x + c*x^2])/(c^2*e^4) + ((8*c*e*f - 8*c*d*g + 3*b*e*g + 6*c*e*g*x)*(a + b*x + c*x^2)^(3/2))/(24*c*e^2) + ((
3*b^4*e^4*g - 128*c^4*d^3*(e*f - d*g) + 192*c^3*d*e*(b*d - a*e)*(e*f - d*g) - 8*b^2*c*e^3*(b*e*f - b*d*g + 3*a
*e*g) + 48*c^2*e^2*(a^2*e^2*g - b^2*d*(e*f - d*g) + 2*a*b*e*(e*f - d*g)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[
a + b*x + c*x^2])])/(128*c^(5/2)*e^5) + ((c*d^2 - b*d*e + a*e^2)^(3/2)*(e*f - d*g)*ArcTanh[(b*d - 2*a*e + (2*c
*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/e^5

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 828

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^
2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(f+g x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx &=\frac {(8 c e f-8 c d g+3 b e g+6 c e g x) \left (a+b x+c x^2\right )^{3/2}}{24 c e^2}-\frac {\int \frac {\left (\frac {1}{2} \left (8 c e (b d-2 a e) f+4 a c d e g-2 b d \left (4 c d-\frac {3 b e}{2}\right ) g\right )+\frac {1}{2} \left (3 b^2 e^2 g+16 c^2 d (e f-d g)-4 c e (2 b e f-2 b d g+3 a e g)\right ) x\right ) \sqrt {a+b x+c x^2}}{d+e x} \, dx}{8 c e^2}\\ &=-\frac {\left (3 b^3 e^3 g-64 c^3 d^2 (e f-d g)+16 c^2 e (5 b d-4 a e) (e f-d g)-4 b c e^2 (2 b e f-2 b d g+3 a e g)+2 c e \left (3 b^2 e^2 g+16 c^2 d (e f-d g)-4 c e (2 b e f-2 b d g+3 a e g)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 e^4}+\frac {(8 c e f-8 c d g+3 b e g+6 c e g x) \left (a+b x+c x^2\right )^{3/2}}{24 c e^2}+\frac {\int \frac {\frac {1}{4} \left (4 c e (b d-2 a e) (8 c e (b d-2 a e) f+4 a c d e g-b d (8 c d-3 b e) g)-d \left (4 b c d-b^2 e-4 a c e\right ) \left (3 b^2 e^2 g+16 c^2 d (e f-d g)-4 c e (2 b e f-2 b d g+3 a e g)\right )\right )+\frac {1}{4} \left (3 b^4 e^4 g-128 c^4 d^3 (e f-d g)+192 c^3 d e (b d-a e) (e f-d g)-8 b^2 c e^3 (b e f-b d g+3 a e g)+48 c^2 e^2 \left (a^2 e^2 g-b^2 d (e f-d g)+2 a b e (e f-d g)\right )\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{32 c^2 e^4}\\ &=-\frac {\left (3 b^3 e^3 g-64 c^3 d^2 (e f-d g)+16 c^2 e (5 b d-4 a e) (e f-d g)-4 b c e^2 (2 b e f-2 b d g+3 a e g)+2 c e \left (3 b^2 e^2 g+16 c^2 d (e f-d g)-4 c e (2 b e f-2 b d g+3 a e g)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 e^4}+\frac {(8 c e f-8 c d g+3 b e g+6 c e g x) \left (a+b x+c x^2\right )^{3/2}}{24 c e^2}+\frac {\left (\left (c d^2-b d e+a e^2\right )^2 (e f-d g)\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^5}+\frac {\left (3 b^4 e^4 g-128 c^4 d^3 (e f-d g)+192 c^3 d e (b d-a e) (e f-d g)-8 b^2 c e^3 (b e f-b d g+3 a e g)+48 c^2 e^2 \left (a^2 e^2 g-b^2 d (e f-d g)+2 a b e (e f-d g)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^2 e^5}\\ &=-\frac {\left (3 b^3 e^3 g-64 c^3 d^2 (e f-d g)+16 c^2 e (5 b d-4 a e) (e f-d g)-4 b c e^2 (2 b e f-2 b d g+3 a e g)+2 c e \left (3 b^2 e^2 g+16 c^2 d (e f-d g)-4 c e (2 b e f-2 b d g+3 a e g)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 e^4}+\frac {(8 c e f-8 c d g+3 b e g+6 c e g x) \left (a+b x+c x^2\right )^{3/2}}{24 c e^2}-\frac {\left (2 \left (c d^2-b d e+a e^2\right )^2 (e f-d g)\right ) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{e^5}+\frac {\left (3 b^4 e^4 g-128 c^4 d^3 (e f-d g)+192 c^3 d e (b d-a e) (e f-d g)-8 b^2 c e^3 (b e f-b d g+3 a e g)+48 c^2 e^2 \left (a^2 e^2 g-b^2 d (e f-d g)+2 a b e (e f-d g)\right )\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{64 c^2 e^5}\\ &=-\frac {\left (3 b^3 e^3 g-64 c^3 d^2 (e f-d g)+16 c^2 e (5 b d-4 a e) (e f-d g)-4 b c e^2 (2 b e f-2 b d g+3 a e g)+2 c e \left (3 b^2 e^2 g+16 c^2 d (e f-d g)-4 c e (2 b e f-2 b d g+3 a e g)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 e^4}+\frac {(8 c e f-8 c d g+3 b e g+6 c e g x) \left (a+b x+c x^2\right )^{3/2}}{24 c e^2}+\frac {\left (3 b^4 e^4 g-128 c^4 d^3 (e f-d g)+192 c^3 d e (b d-a e) (e f-d g)-8 b^2 c e^3 (b e f-b d g+3 a e g)+48 c^2 e^2 \left (a^2 e^2 g-b^2 d (e f-d g)+2 a b e (e f-d g)\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} e^5}+\frac {\left (c d^2-b d e+a e^2\right )^{3/2} (e f-d g) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^5}\\ \end {align*}

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Mathematica [A]
time = 2.70, size = 427, normalized size = 0.97 \begin {gather*} \frac {\frac {2 e \sqrt {a+x (b+c x)} \left (-9 b^3 e^3 g-16 c^3 \left (12 d^3 g-6 d^2 e (2 f+g x)+2 d e^2 x (3 f+2 g x)-e^3 x^2 (4 f+3 g x)\right )+6 b c e^2 (10 a e g+b (4 e f-4 d g+e g x))+8 c^2 e \left (a e (32 e f-32 d g+15 e g x)+b \left (30 d^2 g-2 d e (15 f+7 g x)+e^2 x (14 f+9 g x)\right )\right )\right )}{c^2}-768 \sqrt {-c d^2+b d e-a e^2} \left (c d^2+e (-b d+a e)\right ) (-e f+d g) \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )-\frac {3 \left (3 b^4 e^4 g+128 c^4 d^3 (-e f+d g)-192 c^3 d e (b d-a e) (-e f+d g)-8 b^2 c e^3 (b e f-b d g+3 a e g)+48 c^2 e^2 \left (a^2 e^2 g+2 a b e (e f-d g)+b^2 d (-e f+d g)\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{c^{5/2}}}{384 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x),x]

[Out]

((2*e*Sqrt[a + x*(b + c*x)]*(-9*b^3*e^3*g - 16*c^3*(12*d^3*g - 6*d^2*e*(2*f + g*x) + 2*d*e^2*x*(3*f + 2*g*x) -
e^3*x^2*(4*f + 3*g*x)) + 6*b*c*e^2*(10*a*e*g + b*(4*e*f - 4*d*g + e*g*x)) + 8*c^2*e*(a*e*(32*e*f - 32*d*g + 1
5*e*g*x) + b*(30*d^2*g - 2*d*e*(15*f + 7*g*x) + e^2*x*(14*f + 9*g*x)))))/c^2 - 768*Sqrt[-(c*d^2) + b*d*e - a*e
^2]*(c*d^2 + e*(-(b*d) + a*e))*(-(e*f) + d*g)*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + x*(b + c*x)])/Sqrt[-(c*d^
2) + e*(b*d - a*e)]] - (3*(3*b^4*e^4*g + 128*c^4*d^3*(-(e*f) + d*g) - 192*c^3*d*e*(b*d - a*e)*(-(e*f) + d*g) -
8*b^2*c*e^3*(b*e*f - b*d*g + 3*a*e*g) + 48*c^2*e^2*(a^2*e^2*g + 2*a*b*e*(e*f - d*g) + b^2*d*(-(e*f) + d*g)))*
Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/c^(5/2))/(384*e^5)

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Maple [A]
time = 0.14, size = 742, normalized size = 1.68

method	result	size
default	\(\frac {g \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{e}+\frac {\left (-d g +e f \right ) \left (\frac {\left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{3}+\frac {\left (e b -2 c d \right ) \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )+\frac {e b -2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (e b -2 c d \right )^{2}}{e^{2}}\right ) \ln \left (\frac {\frac {e b -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{2 e}+\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}+\frac {\left (e b -2 c d \right ) \ln \left (\frac {\frac {e b -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\right )}{2 e \sqrt {c}}-\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (e b -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}\right )}{e^{2}}\)	\(742\)
risch	\(\text {Expression too large to display}\)	\(2742\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(c*x^2+b*x+a)^(3/2)/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

g/e*(1/8*(2*c*x+b)/c*(c*x^2+b*x+a)^(3/2)+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)/c*(c*x^2+b*x+a)^(1/2)+1/8*(4*a*c-b^
2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))+(-d*g+e*f)/e^2*(1/3*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e
)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2*(b*e-2*c*d)/e*(1/4*(2*c*(x+d/e)+(b*e-2*c*d)/e)/c*(c*(x+d/e)^2+(b*e-2*c*d)
/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/c^(3/2)*ln((1/2*
(b*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)))+(a*e^2-b*
d*e+c*d^2)/e^2*((c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2*(b*e-2*c*d)/e*ln((1/2*(b
*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e
^2-b*d*e+c*d^2)/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*
e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(c*x^2+b*x+a)^(3/2)/(e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(2*c*d-%e*b>0)', see `assume?`
for more det

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(c*x^2+b*x+a)^(3/2)/(e*x+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (f + g x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(c*x**2+b*x+a)**(3/2)/(e*x+d),x)

[Out]

Integral((f + g*x)*(a + b*x + c*x**2)**(3/2)/(d + e*x), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(c*x^2+b*x+a)^(3/2)/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (f+g\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x),x)

[Out]

int(((f + g*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x), x)

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